Q13/012) Find a four-digit natural number n, such that the last four digits of n2 are same as n.

This can be extended to

Find a k digit natural number n, such that the last k digits of n² are same as n.( k = 4)

Now n^2 –n = 0 mod 10^k

Or n(n-1)  = 0 mod 10^k

Now as n and n-1 consecutive so one of them is divisible by 2^k( any multiple)  and another by 5^k ( odd multiple)

For k = 1 we get n = 5 or 6

For k= 2 we have 5^2 = 25  and 24 is divisible by  2^2 so n = 25

                           Also 3 * 5^2 + 1 = 76 is divisible by 2^2 so n = 76

For k= 3 we have 5^3 = 125  and we need a multiple of 125 +/- 1 divisible by 8 that is (375+ 1) and (625-1) So Ans are 376 and 625

For k= 4 we have 5^4 = 625 so we need multiple of 625 +/- 1 divisible by 16 they are 624( 625-1) and 625 * 15 + 1 = 9376 but 625 is rejected as it is 3 digit number

So for 4 it is 9376

refer to

http://puzzle-a-day.blogspot.in/2011/07/four-digit-number.html

for another soultion